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Math 51, Winter 2010
Math 51A Section
MATH 51A SECTION
Definition 1.
(a) Let f : Dn ⊆ Rn → R be a differentiable function and a is a point in Dn . Then the gradient of f at the
point a is
∂f
∂f
∂f
∇f (a) =
(a),
(a), , . . .
(a) .
∂x1
∂x2
∂xn
(b) The direction derivative of f at a in the direction of v is
f (a + hv) − f (a)
h→0
h
Dv f (a) = lim
where v is a unit vector. It is equal to
∇f (a) · v
if the above limit exits.
Proposition 1. Suppose f : Dn ⊆ Rn → R is differentiable at a point a in Dn . Then
− k∇f (a)k 6 ∇f (a) · v 6 k∇f (a)k
The maximum of ∇f (a) · v is attained if and only if v = c∇f (a) for some positive number c.
The minimum of ∇f (a) · v is attained if and only if v = c∇f (a) for some negative number c.
Proposition 2. ∇f (a) is :
(a) perpendicular to a level set Sc = {x ∈ X | f (x) = c} if a is on this level set.
(b) the equation for the tangent plane of the level set Sc at a is given by
∇f (a) · (x − a) = 0 .
Remark. tangent plane is really a “plane” if f is a function of 3 variables. If f is a function of (x, y), then
Sc is a curve in R2 so the equation above defines the tangent line of Sc at a.
Definition 2. Let f : Dn ⊆ Rn → R be a differentiable function and a is a point in Dn . Then the graph of f is a
hypersurface in Rn+1 given by xn+1 = f (x1 , x2 , . . . , xn ). The the equation for the tangent plane of the graph at
(a, f (a)) is
xn+1 = f (a) + ∇f (a) · (x − a).
Problem 1. A ladybug is crawling on graph paper. She is at the point (3, 7) and notices that if she moves in
the x-direction, the temperature increases at a rate of 3 deg/cm. If she moves in the y-direction, the temperature
decreases at a rate of 2 deg/cm. In what direction should the ladybug move if
(a) She wants to warm up most rapidly?
(b) She wants to cool down most rapidly?
(c) She desires her temperature not to change?
Answer.
(a)
∂T
∂x (3, 7)
= 3, and
∂T
∂y
(3, 7) = −2. So the direction is in the direction of ∇T (3, 7) =
−3
(b) The direction is in the direction of −∇T (3, 7) =
.
2
1
3
.
−2
Math 51, Winter 2010
Math 51A Section
(c) We
a vector v such that Dv T (3, 7) = ∇T (3, 7) · (v/ kvk) = 0, i.e., any
need
vector which is orthogonal to
3
2
. Therefore v can be any scalar multiple (positive or negative) of
. So she should move in the
−2
3
2
same or in the opposite direction of the vector
.
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Problem 2. An ant is moving on the surface x4 + 4y 3 + z 2 = 5/2 where z ≥ 0. Suppose the temperature on the
surface is given by z. She is at the point (1, 0, 1) now and feels too cold. In what direction she should move in
order to warm up most rapidly?
Answer. Since z ≥ 0, z =
∇f (1, 0).
p
5/2 − x4 − 4y 3 = f (x, y). The required direction is the same direction as the vector
∇f (x, y) =
∇f (1, 0) =
−4x3
−12y 2
5/2 − x4 − 4y 3
!T
√
−4 2
√
.
0
3
p
5/2 − x4 − 4y 3
p
!T
.
Problem 3. Let g : R2 → R be a differentiable function such that g(x, y) = x − y, and let f : R → R be a
differentiable function such that f (R) = R. The composition h = f ◦ g : R2 → R is then a differentiable function
and its value at (x, y) is h(x, y) = f (x − y).
(a) Suppose that f is monotone and so f 0 (t) 6= 0 for all t ∈ R, and that f 0 (1) = 2. Compute ∇h(x, y) and find
∇h(1, 0). Use this or otherwise, sketch the level curve which passes through the point (1, 0) and a few level
curves nearby. Sketch also ∇h(1, 0) in the same diagram.
(b) Suppose now that f 0 (1) = 0 and f 0 (t) 6= 0 otherwise. Sketch the level curve which passes through the point
(1, 0) and a few level curves nearby. You should indicate clearly how the level curves in this part differ
from the ones in part (a).
Answer.
(a)
Jh (x, y) = Jf (g(x, y))Jg (x, y) = f 0 (x − y) 1 −1
Jh (1, 0) = f 0 (1) 1 −1 = ∇h(1, 0)T
1
.
∇h(1, 0) = 2
−1
1
From the above calculation, ∇h(x, y) = f 0 (x − y)
. Since f 0 (x − y) 6= 0 for all x and y by assumption,
−1
1
∇h(x, y) is parallel to the vector
at any point (x, y). It means that if we go in the same or in the
−1
1
1
opposite direction orthogonal to
, i.e., the direction of
, at ANY point, h stays constant. Therefore
−1
1
level curves of h are all straight lines of the form x − y = c where c is any real number. The value of h
along the level set x − y = c is h(c).
(b) The level curve which passes through the point (1, 0) has equation x − y = 1. Since along this curve,
T
∇h(x, x − 1) = f 0 (1) 1 −1 = 0, nearby level curves should be more densely packed than those in part
(a).
Problem 4. Let f (x, y) = ex cos y. Find ∇f (x, y), Jf (x, y), ∇f (0, π), Jf (0, π) and the equation for the tangent
plane of z = ex cos y at the point (0, π, −1).
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Math 51, Winter 2010
Math 51A Section
Answer. Let F (x, y, z) = ex cos y −z. Then ∇F (x, y, z) = (ex cos y, −ex sin y, −1) and ∇F (0, π, −1) = (−1, 0, −1).
The equation for the tangent plane is then
∇F (0, π, −1) · (x, y − π, z + 1) = 0
−x − (z + 1) = 0
z = −x − 1
Problem 5. Find the tangent plane and normal line of x2 − 2y 2 + 5xz = 7 at point (−1, 0, − 56 ).
Answer. 8x + 5z = −14.
Problem 6. Let f (x, y, z) = x2 + y 2 − z 2 . Find the tangent plane and normal line of the level set of f at the level
0 at any point (x, y, z) where the level set is differentiable.
Answer. ∇f (x, y, z) = (2x, 2y, −2z) and is zero only at (0, 0, 0). So the level set at level 0 is differentiable
everywhere except at (0, 0, 0). Suppose (a, b, c) 6= (0, 0, 0) is any point on the level set defined by x2 + y 2 − z 2 = 0.
In particular, a2 + b2 − c2 = 0. The tangent plane at this point is then
(2a, 2b, −2c) · (x − a, y − b, z − c) = 0
2a(x − a) + 2b(y − b) − 2c(z − c) = 0.
The normal line is:
   


x
a
2a
y  =  b  + s  2b  .
z
c
−2c
Problem 7. Find the tangent plane and normal line of z = ex + y at the point (1, 1, e1 + 1).
Answer. ex + y − z = 0.
Problem 8. Find the points on the surface 9x2 − 45y 2 + 5z 2 = 45 where the tangent plane is parallel to the plane
x + 5y − 2z = 7.
2
2
2
Answer. If two surfaces are parallel, their normal vectors are colinear.
 Ifwe let f (x, y, z) = 9x − 45y + 5z , then
1
∇f (a) is a normal vector to the tangent plane at a, and ∇f (a) = λ  5 . Then we get 18x = λ, −90y = 5λ, and
−2
10z = −2λ. Plug these equations into 9x2 − 45y 2 + 5z 2 = 45, we can solve λ. Answer: ( 45 , − 45 , − 92 ) and (− 45 , 45 , 92 )
Problem 9. Find equations for the planes tangent to the surface z = x2 − 6x + y 3 that are parallel to the plane
4x − 12y + z = 7.
Answer. Let F (x, y, z) = x2 − 6x + y 3 − z. Then the surface is the level set at level 0 of F . ∇F (x, y, z) =
(2x − 6, 3y 2 , −1) and we want to find the points on the surface where ∇F (x, y, z) = k(4, −12, 1). Solving the
equaitons, we get k = −1, and from this, it is easy to see that x = 1 and y = ±2. Compute z by using
z = x2 − 6x + y 3 , we have two points (1, 2, −3) and (1, −2, 5). Then it is not hard to find planes tangent to the
surface at those points. The equations are
(4, −12, 1) · (x − 1, y − 2, z + 6) = 0,
and the other plane is
(4, −12, 1) · (x − 1, y + 2, z − 5) = 0
Problem 10. Find equations for the planes tangent to the surface z = sin x + y 2 − 2y that are parallel to the plane
x − 5y + z = 1.
Problem 11. Suppose f : Rm → Rn is a linear transformation given by f (x) = Ax, where A is a m × n matrix.
Find Jf (a).
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